- #1

- 107

- 0

1/(cosA+sinA) + 1/(cosA-sinA) = tan2AcosecA

I got as far as LHS= 2cosA/cos^2A - sin^2A.

I cant go any further, please help!

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter brandon26
- Start date

- #1

- 107

- 0

1/(cosA+sinA) + 1/(cosA-sinA) = tan2AcosecA

I got as far as LHS= 2cosA/cos^2A - sin^2A.

I cant go any further, please help!

- #2

arildno

Science Advisor

Homework Helper

Gold Member

Dearly Missed

- 9,970

- 134

Well, how is cos^2A-sin^2A related to cos2A?

- #3

TD

Homework Helper

- 1,022

- 0

- #4

- 107

- 0

arildno said:Well, how is cos^2A-sin^2A related to cos2A?[/QUOT

one side is equal to the other.

- #5

- 107

- 0

I got as far as simplifying the equation to 2cosA / cos2A. What now?

- #6

Tom Mattson

Staff Emeritus

Science Advisor

Gold Member

- 5,500

- 8

brandon26 said:Please help me prove this identity:

1/(cosA+sinA) + 1/(cosA-sinA) = tan2AcosecA

I got as far as LHS= 2cosA/cos^2A - sin^2A.

So far so good. To finish you'll need the following trig identity:

[tex]\tan(2A)=\frac{2\tan(A)}{1-\tan^2(A)}[/tex]

You'll want to try to make the LHS of your identity look like that. To do that you'll need to divide the numerator and denominator of your expression by some other expression, and you need to figure out which one that is.

Hint: Look at the [itex]1[/itex] in the denominator above. What would you have to divide [itex]\cos^2(A)[/itex] by to get a [itex]1[/itex]?

Last edited:

- #7

arildno

Science Advisor

Homework Helper

Gold Member

Dearly Missed

- 9,970

- 134

The simplest way is now to see:

[tex]\frac{2\cos(A)}{\cos(2A)}=1*\frac{2\cos(A)}{\cos(2A)}=\frac{\sin(2A)}{\sin(2A)}*\frac{2\cos(A)}{\cos(2A)}=\frac{2\cos(A)}{\sin(2A)}\tan(2A)[/tex]

I'll leave the last step to you..

[tex]\frac{2\cos(A)}{\cos(2A)}=1*\frac{2\cos(A)}{\cos(2A)}=\frac{\sin(2A)}{\sin(2A)}*\frac{2\cos(A)}{\cos(2A)}=\frac{2\cos(A)}{\sin(2A)}\tan(2A)[/tex]

I'll leave the last step to you..

Last edited:

- #8

TD

Homework Helper

- 1,022

- 0

Once you have this, substitute tan(2a) by sin(2a)/cos(2a) and use the double angle formula on sin(2a). After that, realize that csc(a) = 1/sin(a) and you should be therebrandon26 said:I got as far as simplifying the equation to 2cosA / cos2A. What now?

- #9

- 107

- 0

arildno said:The simplest way is now to see:

[tex]\frac{2\cos(A)}{\cos(2A)}=1*\frac{2\cos(A)}{\cos(2A)}=\frac{\sin(2A)}{\sin(2A)}*\frac{2\cos(A)}{\cos(2A)}=\frac{2\cos(A)}{\sin(2A)}\tan(2A)[/tex]

I'll leave the last step to you..

tan2A (2cosA/sin2A) = tan2A (2cosA/ (2sinAcosA)) = tan2AcosecA.

Safe boys!

Share: